The equation is really simple

$\displaystyle \begin{aligned} 4000 & = 2^5 \cdot 5^3 & \small \color{#3D99F6} \text{Let } x = 2^a5^d, y = 2^b5^e, z = 2^c5^f \\ & = 2^{a + b + c} \cdot 5^{d + e + f} \end{aligned}$

Our problem reduces down to finding how many ordered triplets of positive integers $(a, b, c)$ and $(d, e, f)$ satisfy $a + b + c = 5$ and $d + e + f = 3$ , respectively.

There are $\displaystyle \binom{3 + 5 - 1}{5} = \binom{7}{5} = 21$ ways to choose $5$ times from $3$ objects $a, b, c$ with repetition. Similarly, there are $\displaystyle \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10$ ways to choose $3$ times from $3$ objects $d, e, f$ with repetition. Together, there are $21 \cdot 10 = 210$ ways to choose $2^a \cdot 2^b \cdot 2^c \cdot 5^d \cdot 5^e \cdot 5^f$ and thus $\boxed{210}$ ordered triplets $(x, y, z)$ of positive integers such that $xyz = 4000$ .